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Unformatted text preview: BASIC BUSINESS STATISTICS BASIC BUSINESS STATISTICS Eleventh Edition Berenson Levine Krehbiel Slides Modified by Eugene Li Topic 4 (Part 2) Topic 4 (Part 2) Experimental Designs (Ch. 11)
s s s s An Introduction to Experimental Design Completely Randomized Design (11.1) Randomized Block Design (11.2) Factorial Design (11.3) Slide2 An Introduction to Experimental Design An Introduction to Experimental Design
s s s s Statistical studies can be classified as being either experimental or observational. In an experimental study, one or more factors are controlled so that data can be obtained about how the factors influence the variables of interest. In an observational study, no attempt is made to control the factors. Causeandeffect relationships are easier to establish in experimental studies than in observational studies. Slide3 An Introduction to Experimental Design An Introduction to Experimental Design
s s s s s A factor is a variable that the experimenter has selected for investigation. A treatment is a level of a factor. Experimental units are the objects of interest in the experiment. A completely randomized design is an experimental design in which the treatments are randomly assigned to the experimental units. If the experimental units are heterogeneous, blocking can be used to form homogeneous groups, resulting in a randomized block design. Slide4 An Introduction to Experimental Design An Introduction to Experimental Design
s s Basically, we are still testing the equality of the means; in addition, we can deduce whether the factor has effect or not on the response variable. We use the ANOVA as the data analysis method to obtain the result. Slide5 Completely Randomized Designs Completely Randomized Designs
s s s s s s AmongGroup Estimate of Population Variance WithinGroup Estimate of Population Variance Comparing the Variance Estimates: The F Test The ANOVA Table (same as in previous) Multiple/Pairwise Comparisons (same as in previous part 1) Note: An extension of comparing the means of two independent populations Slide6 AmongGroup Estimate AmongGroup Estimate of Population Variance
s s the among group estimate of σ 2 is referred to as mean square among (MSA) as in part 1. However, for a real designs practitioner, we call it as the mean square due to treatments (MSTr). The formula for MSA is
MSA = n j ( x j − x) 2 ∑
j =1 c c −1 s s The numerator is called the sum of squares among groups (SSA). The denominator c 1 represents the degrees of freedom associated with SSA.
Slide7 WithinGroup Estimate WithinGroup Estimate of Population Variance
s s The second estimate of σ 2, the withingroup estimate, is referred to as the mean square within (MSW). And it is also known as mean square due to error (MSE), where error means the random variation. The formula for MSW is
MSW = ( n j −1) s 2 ∑ j
j= 1 c n −c = ( xij − x j ) 2 ∑∑
j = i= 1 1 c nj n −c s s The numerator is called the sum of squares within groups (SSW). The denominator n c represents the degrees of freedom associated with SSW.
Slide8 ANOVA Table for a ANOVA Table for a Completely Randomized Design Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Among Within Total SSA c 1 SSW n c SST n 1
MSA = SSA c 1
SSW nc
MSA MSW MSW = Slide9 Example: Home Products, Inc. Example: Home Products, Inc.
Home Products, Inc. is considering marketing a long lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. The number of times each car went through the carwash is shown on the next slide. Home Products, Inc. must decide which wax to market. Are the three waxes equally effective?
Slide10 Example: Home Products, Inc. Example: Home Products, Inc. Wax Wax Type 1 Type 2 48 73 54 63 57 66 54 64 62 74 68 26.5 Wax Type 3 51 63 61 54 56 57 24.5
Slide11 Observation 1 2 3 4 5 Sample Mean 55 Sample Variance 26.0 Example: Home Products, Inc.
s Completely Randomized Design • Hypotheses H0: µ 1 = µ 2 = µ 3 ( = µ , a common mean) Ha: Not all the means are equal where: µ 1 = mean number of washes for Type 1 wax µ 2 = mean number of washes for Type 2 wax µ 3 = mean number of washes for Type 3 wax Note: If H0 is rejected, that is at least one of the means is either smaller than or bigger than the rest, that means at least one type of the wax Slide12 lasts either shorter or longer than the other two types. Example: Home Products, Inc.
s Completely Randomized Design • Mean Square Among (among wax types) Since the sample sizes are all equal _ _ _ = x = (x1 + x2 + x3)/3 = (55 + 68 + 57)/3 = 60 SSA = 5(55 60)2 + 5(68 60)2 + 5(57 60)2 = 490 MSA = 490/(3 1) = 245 • Mean Square Within (within wax type) SSW = 4(26.0) + 4(26.5) + 4(24.5) = 308 MSW = 308/(15 3) = 25.667 Slide13 Example: Home Products, Inc.
s Completely Randomized Design • ANOVA Table Source of Sum of Degrees of Mean Variation Squares Freedom Squares F Among 490 2 245 9.55 Within 308 12 25.667 Total 798 14 Slide14 Example: Home Products, Inc.
s Completely Randomized Design • Rejection Rule Assuming α = .05, F.05 = 3.89 (2 d.f. numerator and 12 d.f. denominator). Reject H0 if F > 3.89. • Test Statistic F = MSA/MSW = 245/25.667 = 9.55 • Conclusion Since F = 9.55 > F.05 = 3.89, we reject H0. The mean number of car washes are not the same for all three waxes, i.e. at least one type lasts differently with the other two. That means, three types of wax are not similar. Slide15 Multiple Comparison: TukeyKramer Multiple Comparison: TukeyKramer
s As this example is exactly the same as in part 1 (just change the situation but keep the data), the TukeyKramer test, the steps and the result will be the same as in part 1. Slide16 You may ask……. You may ask……. What’s the difference between completely randomized design and ANOVA? Completely randomized design is a design, but ANOVA is the data analysis method used in a design. We’ll use ANOVA in the coming tow designs too. Slide17 Randomized Block Design Randomized Block Design
The ANOVA Procedure s Computations and Conclusions s Note: 1. An extension of matched pairs. 2. Requirements of oneway ANOVA also apply
s Slide18 Independent samples or blocks?
s s Groups of matched observations are formed into blocks, in order to remove the effects of “unwanted” variability. By doing so we improve the chances of detecting the variability of interest. Slide19 The ANOVA Procedure for a The ANOVA Procedure for a Randomized Block Design
s s s The ANOVA procedure for the randomized block design requires us to partition the sum of squares total (SST) into three groups: sum of squares among groups, sum of squares among blocks, and sum of squares due to error (means random variation). The formula for this partitioning is SST = SSA + SSBL + SSE The total degrees of freedom, n 1, are partitioned such that c 1 degrees of freedom go to groups, r 1 goes to blocks, and (r 1)(c 1) go to the error term. Slide20 ANOVA Table for a ANOVA Table for a Randomized Block Design
Source of Sum of Degrees of Variation Squares Freedom Among Blocks Among Groups Mean Squares F
MSBL MSE MSA MSE SSBL SSA r 1 c 1 SSBL MSBL = r  1 SSA MSA = c 1
SSE (r − 1)(c − 1) Error Total SSE (r 1)(c 1) MSE = SST n – 1 (or rc1) Slide21 Note: 1. Even though we calculate the Fratio for Blocking, 1. Even we don’t pay too much attention to it. we 2. The degrees of freedom for the SST is n1. It can The 1. be rc1 if we have a complete design, which means rc1 equal sample sizes. equal 3. In SGC, you have to tell which variable is the In “Group” and which is the “Block” as the output won’t say among groups or among blocks. won’t Now…..how to get all the sums of squares?
Slide22 Slide The Sum of Squares…….oh! The Sum of Squares…….oh! SSA = r ∑ ( x. j − x)
j =1
r c 2 SSA MSA = c −1 SSBL MSBL = r −1
2 SSBL = c ∑ ( xi. − x)
i =1 2 SSE = ∑ ∑ ( xij − x. j − xi. + x)
j =1 i =1 c r SSE MSE = (r −1)(c −1) SST = ∑ ∑ ( xij − x)
j =1 i = 1 c r 2 Slide23 Example: Eastern Oil Co. Example: Eastern Oil Co.
Eastern Oil has developed three new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the miles per gallon ratings of the three blends is being conducted to determine if the mean ratings are the same for the three blends. Five automobiles have been tested using each of the three gasoline blends and the miles per gallon ratings are shown on the next slide. H0: µ x = µ y = µ z ( = µ , a common mean) Ha: Not all the means are equal (this means the gasoline has an effect on the mpg) Slide24 Example: Eastern Oil Co. Example: Eastern Oil Co. Automobile Type of Gasoline (Groups) Blocks (Block) Blend X Blend Y Blend Z Means 1 31 30 30 30.333 2 30 29 29 29.333 3 29 29 28 28.667 4 33 31 29 31.000 5 26 25 26 25.667 Groups Means 29.8 28.8 28.4
Slide25 Example: Eastern Oil Co. Example: Eastern Oil Co.
s Randomized Block Design • Total sum of squares The overall sample mean is 29. Thus, SST= (3129)2 + (3029)2 + ……+ (2929)2 + (2629)2 = 62 • Mean Square Among Groups (Types of Gasoline) The overall sample mean is 29. Thus, SSA = 5[(29.8 29)2 + (28.8 29)2 + (28.4 29)2] = 5.2 MSA = 5.2/(3 1) = 2.6
Slide26 Example: Eastern Oil Co. Example: Eastern Oil Co.
s Cont’d • Mean Square Among Blocks (Automobiles) SSBL = 3[(30.333 29)2 + . . . + (25.667 29)2] = 51.33 MSBL = 51.33/(5 1) = 12.8 • Mean Square Due to Error SSE = 62 5.2 51.33 = 5.47 MSE = 5.47/[(3 1)(5 1)] = .68 Slide27 ANOVA Table for Eastern Oil Co. ANOVA Table for Eastern Oil Co.
Source of Sum of Degrees of Variation Squares Freedom Among Auto Among Types Of Gasoline Mean Squares F Error Total Slide28 Example: Eastern Oil Co. Example: Eastern Oil Co.
s Randomized Block Design • Rejection Rule Assuming α = .05, F.05 = 4.46 (2 d.f. numerator and 8 d.f. denominator). Reject H0 if F > 4.46. • Test Statistic F = MSA/MSE = 2.6/.68 = 3.82 • Conclusion Since 3.82 < 4.46, we fail to reject H0. We don’t have enough evidence to conclude that the miles per gallon ratings differ for the three gasoline blends, i.e we don’t have enough evidence to conclude the gasoline has significant effect on the mileage.
Slide29 Multiple Comparisons: TukeyKramer
s s Of course, in the Eastern Oil Co. example, after we have done the ANOVA and conclude we don’t have enough evidence to say the gasoline has significant effect on the mileage, actually we don’t need to do the multiple comparison. Therefore, this example is just for the purpose of showing how to do. Slide30 Multiple Comparisons: TukeyKramer Multiple Comparisons: TukeyKramer
s The test procedure: • Find a critical range (CR) as follows: CR = Qα MSE r • Try to compare to the CR for completely randomized design: CR = Qα MSW 2 1 1 + n n j' j • Can you tell what are the difference?
Slide31 TukeyKramer: Eastern Oil Co.
We had three populations (three types of gasoline), c = 3, and we have 5 automobiles, r = 5 (Note: In this example, every type of gasoline has been put into each of five automobile, therefore r is good enough; otherwise, back to nj and nj’), (r1)(c1) = 2*4 = 8 (Do you notice this is the df of error?), MSE = 0.68, the Q has df = 3 and 8, therefore Q.05 = 4.04 CR = Qα
Population Mean Type X Type Y Type Z MSE 0.68 = 4.04 =1.4899 r 5 x j − x j'
all < CR
Slide32 29.8 X vs. Y: 29.8 – 28.8 = 1 28.8 X vs. Z: 29.8 – 28.4 = 1.4 28.4 Y vs. Z: 28.8 – 28.4 = 0.4 One more thing…Estimated Relative Efficiency One more thing…Estimated Relative Efficiency
s Does the blocking result in an increase in precision in comparing the difference groups than by a completely randomized design? We can use this measure, called Estimated Relative Efficiency. • Est. Rel. Efficiency (RE): (r − 1) MSBL + r (c − 1) MSE RE = (rc − 1) MSE Slide33 Est. Rel. Efficiency: Eastern Oil Co. Est. Rel. Efficiency: Eastern Oil Co.
Est. Rel. Efficiency (RE): (r − 1) MSBL + r (c − 1) MSE RE = (rc − 1) MSE (5 − 1)12.8 + 5(3 − 1)0.68 = = 6.09 (15 − 1)0.68
This value means it would take 6.09 times as many observations (15*6.09=92!) in a completely randomized design as compared to this randomized block design in order to have the same precision in comparing the types of gasoline.
Slide34 Models of Fixed and Random Effects
s s Fixed effects • If all possible levels of a factor are included in our analysis we have a fixed effect ANOVA. • The conclusion ofa fixed effect ANOVA applies only to the levels studied. Random effects • If the levels included in our analysis represent a random sample of all the possible levels, we have a randomeffect ANOVA. • The conclusion of the randomeffect ANOVA applies to all the levels (not only those studied). Slide35 Models of Fixed and Random Effects
s s In some ANOVA models the test statistic of the fixed effects case may differ from the test statistic of the random effect case, in particular factorial experiments. Fixed and random effects examples • Fixed effects The Reed Manufacturing Example : All three plants were included • Random effects To determine if there is a difference in the production rate of 50 machines, four machines are randomly selected and there production recorded. Slide36 Factorial Design Factorial Design
s s s s s Valuable designs when simultaneous conclusions about two or more factors are required. “Factorial” is used because the experimental conditions include all possible combination of the factors, e.g. factor A has r levels and factor B has c levels, totally there will be rc treatment combinations. Main effect: As in the past, does this factor alone has effect on the experimental outcomes? Interaction effect: Does the interaction of two factors has effect on the experimental outcomes? Also use ANOVA procedure Slide37 ANOVA Procedure for a Factorial Design ANOVA Procedure for a Factorial Design
s s s The ANOVA procedure for a twofactor (r levels for A and c levels for B) factorial experiment requires us to partition the sum of squares total (SST) into four groups: SS due to factor A, SS due to factor B, SS due to their interaction, and SS due to error. We call this ANOVA as twoway ANOVA. The formula for this partitioning is SST = SSA + SSB+ SSAB + SSE The total degrees of freedom, n – 1 (can be rcn’1 if we have a complete one), are partitioned such that r 1 degrees of freedom go to factor A, c 1 go to factor B, (r 1)(c 1) go to their interaction term, and rc(n’ 1) go to the error term, with n’ = number of replicates.
Slide38 ANOVA Procedure for a Factorial Design ANOVA Procedure for a Factorial Design
s s s For factorial experiments, having “replications” is important when the interaction effect is being investigated. Replication means the number of times each experimental condition is repeated in an experiment. If, say n’, that means each experimental condition is repeated n’ times. If there are r levels for A and c levels for B, and each experimental condition is repeated n’ times; therefore, total number of observations, n , is actually = rcn’. Slide39 ANOVA Procedure for a Factorial Design ANOVA Procedure for a Factorial Design
s The formula for MSA (mean square for factor A) is
MSA = cn' ∑( xi.. − x ) 2
i =1 r r −1 s Use the numerator, which is the SSA and divided by its degree of freedom (r1) will then obtain MSA, as the above. Similarly for factor B, the MSB is s MSB =
s rn' ∑( x. j . − x ) 2
j= 1 c c −1 Use the numerator, which is the SSB and divided by its degree of freedom (c1) will then obtain MSB, as the above.
Slide40 ANOVA Procedure for a Factorial Design ANOVA Procedure for a Factorial Design
s The formula for MSAB (mean square for interaction AB) r c is 2
MSAB = n' ∑ ∑ (xij . − xi.. − x. j . + x)
i =1 j =1 ( r −1)(c −1) s s Use the numerator, which is the SSAB and divided by its degree of freedom (r1)(c1) will then obtain MSAB, as the above. The sum of squares due to error (SSE) can be obtained by SSE = SST SSA SSB SSAB, where SST is SST = ∑ ∑ ∑ ( xijk − x)
i =1 j =1 k = 1 r c n' 2 Slide41 ANOVA Procedure for a Factorial Design ANOVA Procedure for a Factorial Design
s If you prefer, you can find the SSE directly using the formula underneath; however, I strongly feel using SST and minus the rest in getting the SSE is easier. SSE = ∑ ∑ ∑ ( xijk − x ij . ) 2
i =1 j = 1 k =1 r c n' Slide42 ANOVA Table for a Factorial Experiment ANOVA Table for a Factorial Experiment
Source of Sum of Degrees of Mean Variation Squares Freedom Squares F
Factor A Factor B Interaction Error Total SSA SSB r 1 c 1
MSA =
MSB =
MSAB = SSA r 1
SSB c 1 MSA MSE MSB MSE SSAB (r 1)(c 1) SSAB (r − 1)(c − 1) MSAB MSE SSE rc( n’ 1) SST n 1 MSE = SSE rc(n'− 1) Slide43 TwoFactor Analysis of Variance – TwoFactor Analysis of Variance – Factorial Design
s GMAT Example A study involving GMAT score. Scores on the GMAT range from 200 to 800, higher score implying higher aptitude. A university has offered three GMAT preparation program: 1. A 3hour review session covering the types of questions generally asked. 2. A 1day program covering relevant exam material, along with the taking & grading of a sample exam. 3. An intensive 10week course involving the identification of each student’s weaknesses and the setting up of individualized programs for improvement. Slide44 TwoFactor Analysis of Variance – TwoFactor Analysis of Variance – Factorial Design
s GMAT Example –cont’d The GMAT is usually taken by students from three faculties: Business, Engineering and Arts & Science. We have two factors : Program and Faculty and there are 9 treatment combinations. In each treatment combination, we randomly pick two students’ GMAT scores, i.e. two replicates. We have 18 observations (data) in total. Slide45 GMAT Example : The Data GMAT Example : The Data
Business Factor A: Preparation Program 3hr Factor B: Faculty Engineering A&S 540 460(500) 560 620(590) 600 (590) 580 560 Row mean s 500 (540) 580 460 (500) 540 560 (580) 600 450 480 (440) 493.33 400 420 (450) 513.33 480 480 (445) 538.33 410 445 Grand mean= 515
Slide46 1day 10wk Column means GMAT Example: Questions of interest s s s s Are the differences in GMAT scores caused by the different preparation programs? Are the differences in GMAT scores caused by the different faculty? Do students in some faculties do better in GMAT on one type of preparation program whereas others do better on a different type of preparation program? The first two questions are about the main effect of each factor individually to the mean GMAT scores while the last question is about the interaction effect to the mean GMAT scores. Slide47 s s s The experimental designs discussed so far cannot provide answers to these questions together. A new experimental design is needed, which is the factorial design. A twofactor ANOVA with interaction will be used for data analysis. Slide48 GMAT Example GMAT Example
Business Engineering A&S 480 400 420 480 480 410 3hr 1day
10wk 500 580 460 540 560 600 540 460 560 620 600 580 Are the differences in GMAT scores caused Are by the different preparation programs? preparation
Slide49 Twoway ANOVA (two factors)
Test whether mean GMAT score of “3hour”, “1day”, and Test whether mean GMAT score of “3hour”, “1day”, and “10week” differ from one another. H0: µ 3hr= µ 1day = µ 10wk Ha: At least one of them differs Calculations are based on the sum of square for factor A SS) Slide50 GMAT Example GMAT Example Business EngineeringA&S
3hr 500 580 460 540 560 600 540 460 560 620 600 580 480 400 420 480 480 410 1day 10wk Are the differences in GMAT scores caused Are by the different faculty? faculty?
Slide51 Twoway ANOVA (two factors) Test whether mean GMAT scores of the “Business”, “Engineering” and “Arts & Science” differ from one another.
Calculations are based on H0: µ Busi = µ Engg = µ AS the sum of square for factor B Ha: At least one of them SSB differs Slide52 Twoway ANOVA (two factors) Twoway ANOVA (two factors) Business Engineering A&S 500 540 480 3hr 580 460 400 1day 10wk
460 540 560 600 560 620 600 580 420 480 480 410 Are there differences in the mean GMAT score caused by interaction between preparation program and faculty? Slide53 Twoway ANOVA (two factors)
Test whether mean GMAT scores differ due to Test whether mean GMAT scores differ due to interaction. H0: µ 3hr&Busi = µ 3hr&Engg = µ 3hr&AS = µ 1day&Busi =…. = µ 10wk&AS Ha: At least one of them differs
Calculations are based on the sum of square for factors A & SSAB Slide54 Graphical description of the possible relationships between factors A and B. Slide55 M e a n Difference between the levels of factor Difference between the levels of factor A A, and difference between the levels of factor No no B; difference between the levels of factor B interaction R Level 1 of factor B M R Level 1and 2 of factor B e ee s s p ap Level 2 of factor B n o o n n s s e e Levels of factor A Levels of factor A 1 2 3 1 2 Interaction 3 MR MR ee No e e difference between the levels of factor A. s s Difference between the levels of factor B a p ap no n no n s e s e Levels of factor A 1 2 3 1 2 Levels of factor A 3 Slide56 About Interaction plot About Interaction plot
• • • You can observe the vertical axis is the response variable How about the horizontal axis? Which variable (factor) should be put onto it? It doesn’t matter; however, better not to be the one with nominal scale unless both are. Slide57 Requirements: Requirements: (written in the context)
1. 2. 3. The response distributions is normal (i.e. the GMAT scores in each combination of preparation program and faculty is normally distributed) The treatment variances are equal (i.e. the variances of the GMAT scores in each combination of preparation program and faculty are equal) The samples (the GMAT scores of every student) are independent. Slide58 Twoway ANOVA
s GMAT Example – continued • Test of the difference in mean GMAT scores among the three preparation programs H0: µ 3hr= µ 1day = µ 10wk
Ha: At least one of them differs ANOVA Source of Variation Programs Faculties Interaction Error Total SS 6100.0 45300.0 11200.0 19850.0 82450.0 df 2 2 4 9 17 MS 3050.0 22650.0 2800.0 2205.6 F 1.38 10.27 1.27 Pvalue 0.2994 0.0048 0.3503 Factor A Preparation Programs Slide59 F tests for the Twoway ANOVA
s GMAT Example continued Test of the difference in mean GMAT scores among the three preparation programs H0: µ 3hr= µ 1day = µ 10wk
Ha: At least one of them differs F = MS(Programs)/MSE = 1.38 Decision pt.= F.05,2,9 = 4.26; (pvalue = .2994) •Therefore, we don’t have enough evidence to conclude that different preparation program will have different average GMAT scores. Slide60 Twoway ANOVA Twoway ANOVA
s GMAT Example continued Test whether mean GMAT scores of the “Business”, “Engineering” and “Arts & Science” differ from one another. H0: µ Busi = µ Engg = µ AS Ha: At least one of them differs ANOVA Source of Variation Programs Faculties Interaction Error Total SS 6100.0 45300.0 11200.0 19850.0 82450.0 df 2 2 4 9 17 MS 3050.0 22650.0 2800.0 2205.6 F 1.38 10.27 1.27 Pvalue 0.2994 0.0048 0.3503 Factor B Undergraduate faculties
Slide61 F tests for the Twoway ANOVA F tests for the Twoway ANOVA
s GMAT Example continued Test of the difference in mean GMAT scores among the three undergraduate faculties H0: µ Busi= µ Engg = µ AS
Ha: At least one of them differs F = MS(Faculties)/MSE = 10.27 Decision pt.= F.05,2,9 = 4.26; (pvalue = 0.0048) • Therefore, we have enough evidence to conclude that different undergraduate faculty will have different average GMAT scores.
Slide62 Twoway ANOVA Twoway ANOVA
GMAT Example – continued Test whether mean GMAT scores differ due to interaction. H0: µ 3hr&Busi = µ 3hr&Engg = µ 3hr&AS = µ 1day&Busi =…. = µ 10wk&AS s Ha: At least one of them differs
ANOVA Source of Variation Programs Faculties Interaction Error Total SS 6100.0 45300.0 11200.0 19850.0 82450.0 df 2 2 4 9 17 MS 3050.0 22650.0 2800.0 2205.6 F 1.38 10.27 1.27 Pvalue 0.2994 0.0048 0.3503 Interaction AB Programs & Faculties
Slide63 F tests for the Twoway ANOVA F tests for the Twoway ANOVA
s GMAT Example continued Test of the difference in mean GMAT scores due to the interactions H0: µ 3hr&Busi = µ 3hr&Engg = µ 3hr&AS = µ 1day&Busi =…. = µ 10wk&AS Ha: At least one of them differs
F = MS(Interaction)/MSE = 1.27 Decision pt.= F.05,4,9 = 3.63; (pvalue = 0.3503) • Therefore, we don’t have enough evidence to conclude students in some faculties do better in GMAT on one type of preparation program whereas others do better on a different type of preparation program.
Slide64 Example: GMAT Example: GMAT
Several more questions: 1. What is the meaning of the interaction of the factor “Preparation Program” and the factor “Faculty”? 2. Is it possible to have both main effects significant, but interaction effect insignificant? 3. Is it possible to have one or both main effects insignificant, but interaction effect is significant?
s
s Note: So far you have always use MSE as the denominator to obtain all the F ratios for factorial experiment; however, it is not always the true. It is true for the fixed effect models; however for the random effect models, we will use the MS(interaction) instead of MSE as the denominator to calculate the F ratios of the main effects. This will be discussed if you study more statistics about experimental designs in the future. Slide65 Multiple Comparisons: TukeyKramer Multiple Comparisons: TukeyKramer
s Yes, we can also perform the TukeyKramer to Factorial design: • The critical range (CR) for factor A is: CR = Qα MSE cn' where cn’ = number of observations for each level of factor A, and Qα has df r and rc(n’1) • The CR for factor B is: CR = Qα MSE rn' where rn’ = number of observations for each level of factor B, and Qα has df c and rc(n’1) Slide66 TukeyKramer: GMAT example TukeyKramer: GMAT example
s For Program (actually don’t need to do this because the ANOVA result tell us not enough evidence to say at least one differs! Just for demonstration) MSE=2205.6, the Q has df=6 and 9; therefore the critical range (CR) for Program is: CR = Q0.05
Program Mean 3 hrs 1 day 10 wks MSE 2205.6 = 5.02 = 96.2480 cn' (3)(2) 3hrs 493.33 vs. 1day: 493.33 – 513.33 = 20 3hrs 513.33 vs. 10wks: 493.33 – 538.33 = 45 1day vs. 10wks: 513.33 – 538.33 = 25 538.33 all < CR
Slide67 TukeyKramer: GMAT example TukeyKramer: GMAT example
s For Faculty MSE=2205.6, the Q has df=6 and 9; therefore the critical range (CR) for Faculty is (same as for Program because same sample sizes!): CR = Q0.05 MSE 2205.6 = 5.02 = 96.2480 rn' (3)(2) Faculty Busi Engg AS Mean 540 Busi vs. Engg: 540 – 560 = 20 560 Busi vs. AS: 540 – 445 = 95 > 445 Engg vs. AS: 560 – 445 = 115 CR <CR Slide68 Multiple Comparisons: TukeyKramer Multiple Comparisons: TukeyKramer
s Challenge: Textbook didn’t give you the CR for comparing combinations, can you guess? • The critical range (CR) for factor A is: CR = Qα MSE n' where n’ = number of replicates for each combination and Qα has df rc and rc(n’1) (As an exercise, use the above to test the nine combinations. How many pairs you can test?) Slide69 The End of Topic 4 (Part 2) The End of Topic 4 (Part 2) Slide70 ...
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This note was uploaded on 01/17/2010 for the course STATS 2225 taught by Professor Li during the Spring '09 term at Langara.
 Spring '09
 Li
 Statistics

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